Question

30 points Question
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one subQuestion is equal to 10 points
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Answer 1

Question: If $\mathrm{cosA=-\frac{12}{13}}$ and $\mathrm{cotB=\frac{24}{7}}$, where $\mathrm{A}$ lies in the second quadrant and $\mathrm{B}$ in the third quadrant, find the values of the following:

$\small{\mathrm{(i)\:sin(A+B),\:(ii)\:cos(A+B),\:(iii)\:tan(A+B).}}$

Solution:

Step 1.

Here, $\boxed{\mathrm{cosA=-\frac{12}{13}}}$

Remember, $\mathrm{sinA}$ is positive because $\mathrm{A}$ lies in the second quadrant.

$\therefore \mathrm{sinA=\sqrt{1-cos^{2}A}}$

$\Rightarrow \mathrm{sinA=\sqrt{1-\left(-\frac{12}{13}\right)^{2}}}$

$\Rightarrow \boxed{\mathrm{sinA=\frac{5}{13}}}$

Use $\mathrm{tan\theta=\frac{sin\theta}{cos\theta}}$ and remember that $\mathrm{tanA}$ is negative because $\mathrm{A}$ lies in the first quadrant.

Then, $\boxed{\mathrm{tanA=-\frac{5}{12}}}$

Step 2.

Here, $\mathrm{cotB=\frac{24}{7}}$

Then $\boxed{\mathrm{tanB=\frac{7}{24}}}$

Use $\mathrm{sin\theta=\frac{perpendicular}{\sqrt{(perpendicular)^{2}+(base)^{2}}}}$

Remember, both $\mathrm{sinB}$ and $\mathrm{cosB}$ are negative because $\mathrm{B}$ lies in the third quadrant.

$\therefore \mathrm{sinB=-\frac{7}{\sqrt{7^{2}+24^{2}}}}$

$\Rightarrow \boxed{\mathrm{sinB=-\frac{7}{25}}}$

and $\mathrm{cosB=-\sqrt{1-sin^{2}B}}$

$\Rightarrow \mathrm{cosB=-\sqrt{1-\left(-\frac{7}{25}\right)^{2}}}$

$\Rightarrow \boxed{\mathrm{cosB=-\frac{24}{25}}}$

Step 3.

(i)

$\therefore \mathrm{sin(A+B)}$

$=\mathrm{sinA\:cosB+cosA\:sinB}$

$=\mathrm{\frac{5}{13}\:(-\frac{24}{25})+(-\frac{12}{13})\:(-\frac{7}{25})}$

$=\mathrm{-\frac{120}{325}+\frac{84}{325}}$

$=\mathrm{-\frac{36}{325}}$

$\Rightarrow \boxed{\mathrm{sin(A+B)=-\frac{36}{325}}}$

(ii)

$\therefore \mathrm{cos(A+B)}$

$=\mathrm{cosA\:cosB-sinA\:sinB}$

$=\mathrm{(-\frac{12}{13})\:(-\frac{24}{25})-\frac{5}{13}\:(-\frac{7}{25})}$

$=\mathrm{\frac{288}{325}+\frac{35}{325}}$

$=\mathrm{\frac{323}{325}}$

$\Rightarrow \boxed{\mathrm{cos(A+B)=\frac{323}{325}}}$

(iii)

$\therefore \mathrm{tan(A+B)}$

$=\mathrm{\frac{tanA+tanB}{1-tanA\:tanB}}$

$=\mathrm{\frac{-\frac{5}{12}+\frac{7}{24}}{1-(-\frac{5}{12})\:\frac{7}{24}}}$

$=\mathrm{\frac{-\frac{1}{8}}{1+\frac{35}{288}}}$

$=\mathrm{\frac{-\frac{1}{8}}{\frac{323}{288}}}$

$=\mathrm{-\frac{36}{323}}$

$\Rightarrow \boxed{\mathrm{tan(A+B)=-\frac{36}{323}}}$